message stringlengths 2 65.1k | message_type stringclasses 2 values | message_id int64 0 1 | conversation_id int64 0 108k | cluster float64 14 14 | __index_level_0__ int64 0 217k |
|---|---|---|---|---|---|
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Emuskald is addicted to Codeforces, and keeps refreshing the main page not to miss any changes in the "recent actions" list. He likes to read thread conversations where each thread consists of multiple messages.
Recent actions shows a list of n different threads ordered by the time of the latest message in the thread. When a new message is posted in a thread that thread jumps on the top of the list. No two messages of different threads are ever posted at the same time.
Emuskald has just finished reading all his opened threads and refreshes the main page for some more messages to feed his addiction. He notices that no new threads have appeared in the list and at the i-th place in the list there is a thread that was at the ai-th place before the refresh. He doesn't want to waste any time reading old messages so he wants to open only threads with new messages.
Help Emuskald find out the number of threads that surely have new messages. A thread x surely has a new message if there is no such sequence of thread updates (posting messages) that both conditions hold:
1. thread x is not updated (it has no new messages);
2. the list order 1, 2, ..., n changes to a1, a2, ..., an.
Input
The first line of input contains an integer n, the number of threads (1 β€ n β€ 105). The next line contains a list of n space-separated integers a1, a2, ..., an where ai (1 β€ ai β€ n) is the old position of the i-th thread in the new list. It is guaranteed that all of the ai are distinct.
Output
Output a single integer β the number of threads that surely contain a new message.
Examples
Input
5
5 2 1 3 4
Output
2
Input
3
1 2 3
Output
0
Input
4
4 3 2 1
Output
3
Note
In the first test case, threads 2 and 5 are placed before the thread 1, so these threads must contain new messages. Threads 1, 3 and 4 may contain no new messages, if only threads 2 and 5 have new messages.
In the second test case, there may be no new messages at all, since the thread order hasn't changed.
In the third test case, only thread 1 can contain no new messages.
Submitted Solution:
```
def main():
n = int(input())
lista = [int(i) for i in input().split(' ')]
id_start = lista.index(1)
out = 0
org = sorted(lista[id_start:])
for i in range(id_start, n):
if lista[i:] == org[i-id_start:]:
out = i
break
print(i)
if '__main__' == __name__:
main()
``` | instruction | 0 | 6,949 | 14 | 13,898 |
No | output | 1 | 6,949 | 14 | 13,899 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day. | instruction | 0 | 7,025 | 14 | 14,050 |
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
a = input()
b = a.split()
n = int(b[0])
k = int(b[1])
d = int(b[2])
raw = [' ']*n
if n > k**d:
print(-1)
exit()
table = [[1]*d]
p = 1
for i in range(d):
cv = 1
for j in range(n):
if j % p == 0 and j != 0:
cv = cv + 1
if cv == k+1:
cv = 1
raw[j]= str(cv)
print(' '.join(raw))
p = p * k
``` | output | 1 | 7,025 | 14 | 14,051 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day. | instruction | 0 | 7,026 | 14 | 14,052 |
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
import os
import sys
from io import BytesIO, IOBase
import math
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
import os
import sys
from io import BytesIO, IOBase
def main():
pass
# region fastio
BUFSIZE = 8192
class FastIO(IOBase):
newlines = 0
def __init__(self, file):
self._fd = file.fileno()
self.buffer = BytesIO()
self.writable = "x" in file.mode or "r" not in file.mode
self.write = self.buffer.write if self.writable else None
def read(self):
while True:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
if not b:
break
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines = 0
return self.buffer.read()
def readline(self):
while self.newlines == 0:
b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
self.newlines = b.count(b"\n") + (not b)
ptr = self.buffer.tell()
self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
self.newlines -= 1
return self.buffer.readline()
def flush(self):
if self.writable:
os.write(self._fd, self.buffer.getvalue())
self.buffer.truncate(0), self.buffer.seek(0)
class IOWrapper(IOBase):
def __init__(self, file):
self.buffer = FastIO(file)
self.flush = self.buffer.flush
self.writable = self.buffer.writable
self.write = lambda s: self.buffer.write(s.encode("ascii"))
self.read = lambda: self.buffer.read().decode("ascii")
self.readline = lambda: self.buffer.readline().decode("ascii")
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")
n,k,d=map(int,input().split())
if(n>k**d):
print(-1)
else:
ans=[i for i in range(0,n)]
for i in range(0,d):
for j in range(0,n):
print(ans[j]%k+1,end=" ")
ans[j]//=k
print("")
``` | output | 1 | 7,026 | 14 | 14,053 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day. | instruction | 0 | 7,027 | 14 | 14,054 |
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
#http://codeforces.com/problemset/problem/459/C
# Input of the values
n, k, d = map(int, input().split())
#Convention : Busses are numbered from 1 to k
# Students are numbered from 0 to k - 1
# Days are numbered from 0 to k - 1
# Variable for success
success = True
# answer is a 2-d array that stores the bus number each student used on a particular day
# Students are numbered from 0 to n-1
answer = [[] for x in range(n)]
# Initialising the first students details manually
# The first student uses the bus 1 for all of the days
answer[0] = [1 for x in range(d)]
# we will be filling the details for each and every student and
# would print -1 only if we cannot allot any bus number to the next student
for i_th_student in range(1,n):
#answer for previous value
i_minus_one_student = answer[i_th_student - 1]
# Answer to the current student
current_student = list(i_minus_one_student)
# Days are also numbered from 0 to d-1
for i_th_day in reversed(range(d)):
# the condition to change the bus number
if i_minus_one_student[i_th_day] < k :
current_student[i_th_day] += 1
# all the numbers next to it are reset
for i in range(i_th_day + 1, d):
current_student[i] = 1
break
#Save the value
answer[i_th_student] = current_student
#Failed
if current_student == i_minus_one_student:
success = False
break
# printing the output
if not(success):
print ("-1")
else:
# ans_trans is used for giving the output fast
answer_trans= [[] for i in range(d)]
for y in range(n):
i = 0
for x in answer[y]:
answer_trans[i].append(x)
i += 1
for bla in answer_trans:
print (' '.join(map(str, bla)))
``` | output | 1 | 7,027 | 14 | 14,055 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day. | instruction | 0 | 7,028 | 14 | 14,056 |
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
import time,math,bisect,sys
sys.setrecursionlimit(100000)
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IO(): # to take string input
return stdin.readline()
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def readTree(): # to read tree
v=int(input())
adj=[set() for i in range(v+1)]
for i in range(v-1):
u1,u2=In()
adj[u1].add(u2)
adj[u2].add(u1)
return adj,v
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def sieve():
li=[True]*1000001
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime=[]
for i in range(1000001):
if li[i]==True:
prime.append(i)
return prime
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
#####################################################################################
mx=10**9+7
def solve():
n,k,d=IP()
if n>pow(k,d):
print(-1)
return
else:
li=[[0 for j in range(n)] for i in range(d)]
student=0
for i in range(1,n):
for j in range(d):
li[j][i]=li[j][i-1]
for j in range(d):
li[j][i]=(li[j][i]+1)%k
if li[j][i]:
break
for i in range(d):
for j in range(n):
print(int(li[i][j])+1,end=" ")
print()
return
#t=II()
#for i in range(t):
solve()
#######
#
#
####### # # # #### # # #
# # # # # # # # # # #
# #### # # #### #### # #
###### # # #### # # # # #
``` | output | 1 | 7,028 | 14 | 14,057 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day. | instruction | 0 | 7,029 | 14 | 14,058 |
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
from sys import stdin, stdout, setrecursionlimit
input = stdin.readline
# import string
# characters = string.ascii_lowercase
# digits = string.digits
# setrecursionlimit(int(1e6))
# dir = [-1,0,1,0,-1]
# moves = 'NESW'
inf = float('inf')
from functools import cmp_to_key
from collections import defaultdict as dd
from collections import Counter, deque
from heapq import *
import math
from math import floor, ceil, sqrt
def geti(): return map(int, input().strip().split())
def getl(): return list(map(int, input().strip().split()))
def getis(): return map(str, input().strip().split())
def getls(): return list(map(str, input().strip().split()))
def gets(): return input().strip()
def geta(): return int(input())
def print_s(s): stdout.write(s+'\n')
def solve():
n, k, d = geti()
num = 1
for i in range(1, d+1):
num *= k
if num >= n:
break
else:
print(-1)
return
def convert(prev):
ans = prev[:]
i = d - 1
ans[i] += 1
while ans[i] > k and i > 0:
ans[i-1] += 1
ans[i] = 1
i -= 1
return ans
ans = [[1]*d]
for i in range(1, n+1):
ans.append(convert(ans[i-1]))
# print(ans)
for i in range(d):
print(*[ans[j][i] for j in range(n)])
# 1 2 1 2 1 2 2
# 2 1 2 1 1 2 2
# 1 2 2 1 1 2 1
#
#
# 2 1 1 1 1 2 1 1
# 1 2 1 1 1 2 2 1
# 1 1 2 1 1 1 2 1
# 1 1 1 2 1 1 1 1
# 1 1 1 1 2 1 1 1
if __name__=='__main__':
solve()
``` | output | 1 | 7,029 | 14 | 14,059 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day. | instruction | 0 | 7,030 | 14 | 14,060 |
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
def f(t):
i = -1
t[i] += 1
while t[i] > k:
t[i] = 1
i -= 1
t[i] += 1
return list(map(str, t))
n, k, d = map(int, input().split())
if k ** d < n: print(-1)
else:
t = [1] * d
t[-1] = 0
s = [f(t) for i in range(n)]
print('\n'.join([' '.join(t) for t in zip(*s)]))
``` | output | 1 | 7,030 | 14 | 14,061 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day. | instruction | 0 | 7,031 | 14 | 14,062 |
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
n,k,d = map(int, input().split())
if (n > k ** d):
print(-1)
exit()
for i in range(d):
power = k**i
resArray = [((j // power) % k +1) for j in range(n)]
#for j in range(n//(k**i)+1):
# resArray.extend([(j % k) + 1] * (k ** i))
print(' '.join(map(str,resArray)))
``` | output | 1 | 7,031 | 14 | 14,063 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day. | instruction | 0 | 7,032 | 14 | 14,064 |
Tags: combinatorics, constructive algorithms, math
Correct Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
from typing import List
"""
created by shhuan at 2020/1/8 21:47
"""
def bit(val, n, l):
x = []
while val > 0:
x.append(val % n)
val //= n
return [0] * (l - len(x)) + x[::-1]
def solve(N, K, D):
if math.log(N, K) > D or K ** D < N:
print(-1)
return
a = []
for i in range(N):
x = bit(i, K, D)
a.append(x)
ans = []
for d in range(D):
ans.append(' '.join(map(str, [a[i][d] + 1 for i in range(N)])))
print('\n'.join(ans))
N, K, D = map(int, input().split())
solve(N, K, D)
``` | output | 1 | 7,032 | 14 | 14,065 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
Submitted Solution:
```
#!/usr/bin/python3
import sys
n, k, d = list(map(int, sys.stdin.readline().split()))
x = 1
while x ** d < n:
x += 1
if x > k:
sys.stdout.write("-1\n")
sys.exit(0)
ans = [[1 for i in range(d)] for j in range(n)]
for i in range(1, n):
for j in range(d):
ans[i][j] = ans[i - 1][j]
ans[i][d - 1] += 1
memo = 0
for j in range(d - 1, -1, -1):
ans[i][j] += memo
memo = 0
if ans[i][j] > x:
memo = ans[i][j] - x
ans[i][j] = 1
for i in range(d):
sys.stdout.write(' '.join([str(ans[j][i]) for j in range(n)]) + '\n')
``` | instruction | 0 | 7,033 | 14 | 14,066 |
Yes | output | 1 | 7,033 | 14 | 14,067 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
Submitted Solution:
```
n, k, d = map(int, input().split())
bits = n
for i in range(d):
bits = bits//k + (bits % k != 0)
if bits > 1:
print("-1")
exit()
for _ in range(d):
i = 0
res = 1
while i < n:
for j in range(bits):
print(res, end=' ')
i += 1
if i == n:
break
res += 1
if res > k:
res = 1
bits *= k
print()
``` | instruction | 0 | 7,034 | 14 | 14,068 |
Yes | output | 1 | 7,034 | 14 | 14,069 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
Submitted Solution:
```
#!/usr/bin/python3
import sys
# 1β<=βn,βdβ<=β1000, 1β<=βkβ<=β10**9
n, k, d = map(int, sys.stdin.readline().split())
no_sol = False
solution = [[1 for j in range(n)] for i in range(d)]
def schedule(i, j, level):
global no_sol
if level >= d:
no_sol = True
return
count = j - i
chunk = count // k
extra = count % k
r = i
for t in range(min(k, count)):
size = chunk + (1 if t < extra else 0)
for z in range(size):
solution[level][r+z] = t+1
if size > 1:
schedule(r, r + size, level + 1)
r += size
if k == 1:
if n > 1:
no_sol = True
else:
schedule(0, n, 0)
if no_sol:
print(-1)
else:
for l in solution:
print(' '.join(str(x) for x in l))
``` | instruction | 0 | 7,035 | 14 | 14,070 |
Yes | output | 1 | 7,035 | 14 | 14,071 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
Submitted Solution:
```
import time,math,bisect,sys
sys.setrecursionlimit(100000)
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IO(): # to take string input
return stdin.readline()
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def readTree(): # to read tree
v=int(input())
adj=[set() for i in range(v+1)]
for i in range(v-1):
u1,u2=In()
adj[u1].add(u2)
adj[u2].add(u1)
return adj,v
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def sieve():
li=[True]*1000001
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime=[]
for i in range(1000001):
if li[i]==True:
prime.append(i)
return prime
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
#####################################################################################
mx=10**9+7
def solve():
n,k,d=IP()
if n>pow(k,d):
print(-1)
return
elif d==1:
li=[i for i in range(1,n+1)]
print(*li)
else:
li=[[0 for j in range(n)] for i in range(d)]
student=0
for i in range(1,n):
for j in range(d):
li[j][i]=li[j][i-1]
for j in range(d):
li[j][i]=(li[j][i]+1)%k
if li[j][i]:
break
for i in range(d):
for j in range(n):
print(int(li[i][j])+1,end=" ")
print()
return
#t=II()
#for i in range(t):
solve()
#######
#
#
####### # # # #### # # #
# # # # # # # # # # #
# #### # # #### #### # #
###### # # #### # # # # #
``` | instruction | 0 | 7,036 | 14 | 14,072 |
Yes | output | 1 | 7,036 | 14 | 14,073 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
Submitted Solution:
```
import sys
input=sys.stdin.readline
n,k,d=map(int,input().split())
if k==1:
print(-1)
exit()
size=n
for i in range(d):
size=size//k+(size%k!=0)
if size>1:
print(-1)
else:
x=1
for i in range(d):
ans=[1+i//x%k for i in range(n)]
print(*ans)
x*=k
``` | instruction | 0 | 7,037 | 14 | 14,074 |
No | output | 1 | 7,037 | 14 | 14,075 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
Submitted Solution:
```
# -*- coding: utf-8 -*-
import math
import collections
import bisect
import heapq
import time
import random
import itertools
import sys
from typing import List
"""
created by shhuan at 2020/1/8 21:47
"""
def bit(val, n, l):
x = []
while val > 0:
x.append(val % n)
val //= n
return [0] * (l - len(x)) + x[::-1]
def solve(N, K, D):
if math.log(N, K) > D or K ** D < N:
print(-1)
return
a = []
for i in range(1, N+1):
x = bit(i, K, D)
a.append(x)
ans = []
for d in range(D):
ans.append(' '.join(map(str, [a[i][d] + 1 for i in range(N)])))
print('\n'.join(ans))
N, K, D = map(int, input().split())
solve(N, K, D)
``` | instruction | 0 | 7,038 | 14 | 14,076 |
No | output | 1 | 7,038 | 14 | 14,077 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
Submitted Solution:
```
m=list(input().split())
n=int(m[0])
k=int(m[1])
d=int(m[2])
if n==1:
for i in range(d):
print("1")
exit()
if k==1:
print("-1")
exit()
if k>=n:
m=list(i for i in range(n))
for i in range(d):
#print(m)
o=str(m[0])
for j in range(1,n):
o+=" "+str(m[j])
print(o)
exit()
if d==1:
print('-1')
exit()
if d>=n:
m=list("1" for i in range(n))
for i in range(n):
m[i]="2"
#print(m)
o=str(m[0])
for j in range(1,n):
o+=" "+str(m[j])
print(o)
#m[i]="1"
for i in range(n,d):
#print(m)
print(o)
exit()
from math import ceil, floor
tc=ceil(n/k) #days
tf=floor(n/k) #len k
if d<tf:
print('-1')
exit()
sc=1
for stp in range(1,d+1):
o=""
kc=0
while kc<n:
for kk in range(1,k+1):
for i in range (sc):
if kc<n:
o+=' '+str(kk)
kc+=1
else:
break
sc*=2
print(o[1:])
``` | instruction | 0 | 7,039 | 14 | 14,078 |
No | output | 1 | 7,039 | 14 | 14,079 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Recently Pashmak has been employed in a transportation company. The company has k buses and has a contract with a school which has n students. The school planned to take the students to d different places for d days (each day in one place). Each day the company provides all the buses for the trip. Pashmak has to arrange the students in the buses. He wants to arrange the students in a way that no two students become close friends. In his ridiculous idea, two students will become close friends if and only if they are in the same buses for all d days.
Please help Pashmak with his weird idea. Assume that each bus has an unlimited capacity.
Input
The first line of input contains three space-separated integers n, k, d (1 β€ n, d β€ 1000; 1 β€ k β€ 109).
Output
If there is no valid arrangement just print -1. Otherwise print d lines, in each of them print n integers. The j-th integer of the i-th line shows which bus the j-th student has to take on the i-th day. You can assume that the buses are numbered from 1 to k.
Examples
Input
3 2 2
Output
1 1 2
1 2 1
Input
3 2 1
Output
-1
Note
Note that two students become close friends only if they share a bus each day. But the bus they share can differ from day to day.
Submitted Solution:
```
import time,math,bisect,sys
sys.setrecursionlimit(100000)
from sys import stdin,stdout
from collections import deque
from fractions import Fraction
from collections import Counter
from collections import OrderedDict
pi=3.14159265358979323846264338327950
def II(): # to take integer input
return int(stdin.readline())
def IO(): # to take string input
return stdin.readline()
def IP(): # to take tuple as input
return map(int,stdin.readline().split())
def L(): # to take list as input
return list(map(int,stdin.readline().split()))
def P(x): # to print integer,list,string etc..
return stdout.write(str(x)+"\n")
def PI(x,y): # to print tuple separatedly
return stdout.write(str(x)+" "+str(y)+"\n")
def lcm(a,b): # to calculate lcm
return (a*b)//gcd(a,b)
def gcd(a,b): # to calculate gcd
if a==0:
return b
elif b==0:
return a
if a>b:
return gcd(a%b,b)
else:
return gcd(a,b%a)
def readTree(): # to read tree
v=int(input())
adj=[set() for i in range(v+1)]
for i in range(v-1):
u1,u2=In()
adj[u1].add(u2)
adj[u2].add(u1)
return adj,v
def bfs(adj,v): # a schema of bfs
visited=[False]*(v+1)
q=deque()
while q:
pass
def sieve():
li=[True]*1000001
li[0],li[1]=False,False
for i in range(2,len(li),1):
if li[i]==True:
for j in range(i*i,len(li),i):
li[j]=False
prime=[]
for i in range(1000001):
if li[i]==True:
prime.append(i)
return prime
def setBit(n):
count=0
while n!=0:
n=n&(n-1)
count+=1
return count
#####################################################################################
mx=10**9+7
def solve():
n,k,d=IP()
if n>k**d:
print(-1)
return
else:
li=[[0 for j in range(n)] for i in range(d)]
student=0
for i in range(n):
x=bin(i)[2::].rjust(d,'0')
for j in range(d):
li[j][student]=x[j]
student+=1
for i in range(d):
for j in range(n):
print(int(li[i][j])+1,end=" ")
print()
return
#t=II()
#for i in range(t):
solve()
#######
#
#
####### # # # #### # # #
# # # # # # # # # # #
# #### # # #### #### # #
###### # # #### # # # # #
``` | instruction | 0 | 7,040 | 14 | 14,080 |
No | output | 1 | 7,040 | 14 | 14,081 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends. | instruction | 0 | 7,095 | 14 | 14,190 |
Tags: binary search, sortings, two pointers
Correct Solution:
```
def kac(l,d):
l.sort(reverse=True)
ma=l[0][1]
v=l[0][1]
i=j=0
for i in range(1,len(l)):
v=v+l[i][1]
while abs(l[i][0]-l[j][0])>=d:
v-=l[j][1]
j+=1
if(v>ma):
ma=v
print(ma)
return
n,d=map(int,input().split())
l1=[]
for i in range(n):
m=list(map(int,input().split()))
l1.append(m)
kac(l1,d)
``` | output | 1 | 7,095 | 14 | 14,191 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends. | instruction | 0 | 7,096 | 14 | 14,192 |
Tags: binary search, sortings, two pointers
Correct Solution:
```
n, d = map(int, input().split())
friends = []
for i in range(n):
friends.append(tuple(map(int, input().split())))
friends.sort(key=lambda x: x[0])
prefix = [0] * n
prefix[0] = friends[0][1]
for i in range(1, n):
prefix[i] += prefix[i - 1]
le = 0
r = 0
curr_sum = 0
ans = max(f[1] for f in friends)
while le < n:
while r < n and abs(friends[r][0] - friends[le][0]) < d:
curr_sum += friends[r][1]
r += 1
ans = max(ans, curr_sum)
curr_sum -= friends[le][1]
le += 1
print(ans)
``` | output | 1 | 7,096 | 14 | 14,193 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends. | instruction | 0 | 7,097 | 14 | 14,194 |
Tags: binary search, sortings, two pointers
Correct Solution:
```
[friendsSize, moneyDiff] = [int(x) for x in input().split()]
friends = []
for i in range(friendsSize):
friendMoney, currentPoints = map(int, input().split())
friends.append([friendMoney, currentPoints])
friends = sorted(friends)
currentPoints = 0
maxPoints = 0
j = 0
for i in range(friendsSize):
while j < friendsSize and abs(friends[i][0] - friends[j][0]) < moneyDiff:
currentPoints += friends[j][1]
j += 1
if currentPoints >= maxPoints:
maxPoints = currentPoints
currentPoints -= friends[i][1]
print(maxPoints)
``` | output | 1 | 7,097 | 14 | 14,195 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends. | instruction | 0 | 7,098 | 14 | 14,196 |
Tags: binary search, sortings, two pointers
Correct Solution:
```
n,d=map(int,input().split())
a=[]
for i in range(n):
a.append((list(map(int,input().split()))))
a.sort()
x=0
y=0
m=0
f=0
while y<n:
if a[y][0]-a[x][0]<d:
m+=a[y][1]
y+=1
f=max(m,f)
else:
m-=a[x][1]
x+=1
print(f)
``` | output | 1 | 7,098 | 14 | 14,197 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends. | instruction | 0 | 7,099 | 14 | 14,198 |
Tags: binary search, sortings, two pointers
Correct Solution:
```
user_input=input()
first_line=user_input.split(' ')
n=int(first_line[0])
d=int(first_line[1])
long_list=[]
for i in range(n):
extended_list=[]
user_input = input()
line = user_input.split(' ')
extended_list.extend([int(line[0]),int(line[1])])
long_list.append(extended_list)
long_list.sort()
index=0
maximum=0
for i in range(len(long_list)):
if maximum<long_list[i][1]:
maximum=long_list[i][1]
answer=maximum
maximum=0
i=0
count=0
looped="false"
while i in range(len(long_list)):
if long_list[i][0]-long_list[index][0]>d:
if count>0 and looped=="false":
i-=count
count=0
index=i
looped="true"
else:
index=i
maximum=0
if long_list[i][0]-long_list[index][0]<d:
count+=1
if maximum==0:
maximum=long_list[index][1]
else:
maximum+=long_list[i][1]
answer=max(answer,maximum)
i+=1
print(answer)
``` | output | 1 | 7,099 | 14 | 14,199 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends. | instruction | 0 | 7,100 | 14 | 14,200 |
Tags: binary search, sortings, two pointers
Correct Solution:
```
def find_company(friends, d):
def mergesort_tuples(tuples, index):
def merge(A, B):
i, j = 0, 0
result = []
while i < len(A) and j < len(B):
if A[i][index] < B[j][index]:
result.append(A[i])
i += 1
else:
result.append(B[j])
j += 1
result += A[i:]
result += B[j:]
return result
def divide(tuples):
if len(tuples) > 1:
middle = len(tuples) // 2
return merge(divide(tuples[:middle]), divide(tuples[middle:]))
return tuples
return divide(tuples)
def solve(friends, d):
friends = mergesort_tuples(friends, 0)
left_ptr = 0
right_ptr = 0
current_friendship = 0
max_friendship = 0
while right_ptr < len(friends):
if friends[right_ptr][0] - friends[left_ptr][0] < d:
current_friendship += friends[right_ptr][1]
right_ptr += 1
else:
max_friendship = max(current_friendship, max_friendship)
current_friendship -= friends[left_ptr][1]
left_ptr += 1
max_friendship = max(current_friendship, max_friendship)
return max_friendship
return solve(friends, d)
if __name__ == "__main__":
n, d = [int(x) for x in input().split()]
friends = []
for _ in range(n):
friends.append(tuple([int(x) for x in input().split()]))
print(find_company(friends, d))
``` | output | 1 | 7,100 | 14 | 14,201 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends. | instruction | 0 | 7,101 | 14 | 14,202 |
Tags: binary search, sortings, two pointers
Correct Solution:
```
z,zz=input,lambda:list(map(int,z().split()))
zzz=lambda:[int(i) for i in stdin.readline().split()]
szz,graph,mod,szzz=lambda:sorted(zz()),{},10**9+7,lambda:sorted(zzz())
from string import *
from re import *
from collections import *
from queue import *
from sys import *
from collections import *
from math import *
from heapq import *
from itertools import *
from bisect import *
from collections import Counter as cc
from math import factorial as f
from bisect import bisect as bs
from bisect import bisect_left as bsl
from itertools import accumulate as ac
def lcd(xnum1,xnum2):return (xnum1*xnum2//gcd(xnum1,xnum2))
def prime(x):
p=ceil(x**.5)+1
for i in range(2,p):
if (x%i==0 and x!=2) or x==0:return 0
return 1
def dfs(u,visit,graph):
visit[u]=True
for i in graph[u]:
if not visit[i]:
dfs(i,visit,graph)
###########################---Test-Case---#################################
"""
"""
###########################---START-CODING---##############################
n,m=zz()
lst=[]
for _ in range(n):
lst.append(zz())
lst=sorted(lst)
ans=0
l=0
s=0
for r in range(n):
while lst[r][0]-lst[l][0]>=m:
s-=lst[l][1]
l+=1
s+=lst[r][1]
ans=max(ans,s)
print(ans)
``` | output | 1 | 7,101 | 14 | 14,203 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends. | instruction | 0 | 7,102 | 14 | 14,204 |
Tags: binary search, sortings, two pointers
Correct Solution:
```
from bisect import *
n,d = map(int,input().split())
tp = list()
for i in range(n):
m,s = map(int,input().split())
tp.append((m,s))
tp.sort(key = lambda x : x[0])
m = list()
s = list()
for i in range(n):
m.append(tp[i][0])
s.append(tp[i][1])
ps = list()
sn = 0
for i in range(n):
sn += s[i]
ps.append(sn)
mx = -1000
for i in range(n):
ind = bisect_left(m,m[i]+d)
if(i!=0):
mx = max(mx,ps[ind-1]-ps[i-1])
else:
mx = max(mx,ps[ind-1])
print(mx)
``` | output | 1 | 7,102 | 14 | 14,205 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends.
Submitted Solution:
```
n,d = map(int,input().split())
l = []
for i in range(n):
a,b = map(int,input().split())
l.append([a,b])
l.sort()
ans = l[0][1]
c = l[0][1]
i = 0
j = 1
while j<n:
if l[j][0]-l[i][0]<d:
c+=l[j][1]
j+=1
else:
c-=l[i][1]
i+=1
ans = max(ans,c)
print(ans)
``` | instruction | 0 | 7,103 | 14 | 14,206 |
Yes | output | 1 | 7,103 | 14 | 14,207 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends.
Submitted Solution:
```
n,d=map(int,input().split())
l=[]
for i in range(n):
x,y=map(int,input().split())
l.append([x,y])
l.sort()
ans=0
a=[]
i=0
j=0
while i<n:
if l[i][0]-l[j][0]>=d:
a.append(ans)
ans-=l[j][1]
j+=1
else:
ans+=l[i][1]
a.append(ans)
i+=1
print(max(a))
``` | instruction | 0 | 7,104 | 14 | 14,208 |
Yes | output | 1 | 7,104 | 14 | 14,209 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends.
Submitted Solution:
```
def max_friendship_factor(friends, min_money_diff):
sorted_friends = sorted(friends, key=lambda t: t[0])
n = len(friends)
_, ff_sum = sorted_friends[0] # set first elements ff to ff_sum
max_ff_sum = ff_sum # and max_ff_sum
head_index = 0
hm, hff = sorted_friends[0] # set head money and head ff
for i in range(1, n):
m, ff = sorted_friends[i]
ff_sum += ff
while abs(m - hm) >= min_money_diff:
# push head forwards
ff_sum -= hff
head_index += 1
hm, hff = sorted_friends[head_index]
if max_ff_sum < ff_sum:
max_ff_sum = ff_sum
return max_ff_sum
def run_alg():
first_line_data = input().split(' ')
num_friends = int(first_line_data[0])
min_money_diff = int(first_line_data[1])
friends = []
for _ in range(num_friends):
line_data = input().split(' ')
money = int(line_data[0])
friendship = int(line_data[1])
friends.append((money, friendship))
print(max_friendship_factor(friends, min_money_diff))
if __name__ == '__main__':
run_alg()
``` | instruction | 0 | 7,106 | 14 | 14,212 |
Yes | output | 1 | 7,106 | 14 | 14,213 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends.
Submitted Solution:
```
n,d=map(int,input().split())
l=[]
for i in range(n):
j=tuple(map(int,input().split()))
l.append(j)
k=0
m=0
for i in range(n):
c=l[i][1]
for j in range(n):
if(l[j][0]<l[i][0]+d-1 and l[j][0]+d-1>l[i][0] and i!=j):
c+=l[j][1]
#print(l[j][0]+d,l[i][0])
#print("\n")
if(c>m):
m=c
k=i
print(m)
``` | instruction | 0 | 7,108 | 14 | 14,216 |
No | output | 1 | 7,108 | 14 | 14,217 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.
Kefa has n friends, each friend will agree to go to the restaurant if Kefa asks. Each friend is characterized by the amount of money he has and the friendship factor in respect to Kefa. The parrot doesn't want any friend to feel poor compared to somebody else in the company (Kefa doesn't count). A friend feels poor if in the company there is someone who has at least d units of money more than he does. Also, Kefa wants the total friendship factor of the members of the company to be maximum. Help him invite an optimal company!
Input
The first line of the input contains two space-separated integers, n and d (1 β€ n β€ 105, <image>) β the number of Kefa's friends and the minimum difference between the amount of money in order to feel poor, respectively.
Next n lines contain the descriptions of Kefa's friends, the (i + 1)-th line contains the description of the i-th friend of type mi, si (0 β€ mi, si β€ 109) β the amount of money and the friendship factor, respectively.
Output
Print the maximum total friendship factir that can be reached.
Examples
Input
4 5
75 5
0 100
150 20
75 1
Output
100
Input
5 100
0 7
11 32
99 10
46 8
87 54
Output
111
Note
In the first sample test the most profitable strategy is to form a company from only the second friend. At all other variants the total degree of friendship will be worse.
In the second sample test we can take all the friends.
Submitted Solution:
```
n, d = map(int, input().split())
T = [list(map(int, input().split())) for _ in range(n)]
T.sort(key=lambda x: x[0])
#print(T)
if n == 1:
print(T[0][1])
exit(0)
ans = []
for i in range(len(T) - 1):
temp = T[i][1]
for j in range(i + 1, len(T)):
if T[j][0] - T[i][0] <= d:
temp += T[j][1]
ans.append(temp)
#print(ans)
ans.append(T[-1][1])
print(max(ans))
``` | instruction | 0 | 7,110 | 14 | 14,220 |
No | output | 1 | 7,110 | 14 | 14,221 |
Provide a correct Python 3 solution for this coding contest problem.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5 | instruction | 0 | 7,475 | 14 | 14,950 |
"Correct Solution:
```
N = int(input())
A = [0] * 101010
for i in range(N):
a, b = map(int, input().split())
A[a] += 1
A[b+1] -= 1
ans = 1
cuma = A[0]
for i, a in enumerate(A[1:], 1):
cuma += a
if cuma >= i-1:
ans = i
print(ans-1)
``` | output | 1 | 7,475 | 14 | 14,951 |
Provide a correct Python 3 solution for this coding contest problem.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5 | instruction | 0 | 7,476 | 14 | 14,952 |
"Correct Solution:
```
import itertools
n=int(input())
table=[0 for i in range(100003)]
table[0]=1
for i in range(n):
a,b=map(int,input().split())
table[a]+=1
table[b+1]-=1
table=list(itertools.accumulate(table))
for i in range(len(table)):
if table[i]>=i:
ans=i
print(ans-1)
``` | output | 1 | 7,476 | 14 | 14,953 |
Provide a correct Python 3 solution for this coding contest problem.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5 | instruction | 0 | 7,477 | 14 | 14,954 |
"Correct Solution:
```
n=int(input())
l=[0 for i in range(100003)]
for i in range(n):
a,b=map(int,input().split())
l[a]+=1
l[b+1]-=1
su=1
res=1
for i in range(100003):
su+=l[i]
if su>=i:
res=max(i,res)
print(res-1)
``` | output | 1 | 7,477 | 14 | 14,955 |
Provide a correct Python 3 solution for this coding contest problem.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5 | instruction | 0 | 7,478 | 14 | 14,956 |
"Correct Solution:
```
from itertools import accumulate
n = int(input())
lst = [0] * 100002
for _ in range(n):
a, b = map(int, input().split())
lst[a - 1] += 1
lst[b] -= 1
lst = list(accumulate(lst))
for i in range(n, -1, -1):
if i <= lst[i]:
print(i)
break
``` | output | 1 | 7,478 | 14 | 14,957 |
Provide a correct Python 3 solution for this coding contest problem.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5 | instruction | 0 | 7,479 | 14 | 14,958 |
"Correct Solution:
```
def main():
N = int(input())
ab = [list(map(int, input().split())) for i in range(N)]
imos = [0]*(N+3)
for a, b in ab:
imos[min(a,N+2)] += 1
imos[min(b+1,N+2)] -= 1
for i in range(N+1):
imos[i+1] += imos[i]
#print(imos)
ans = 0
for i, imo in enumerate(imos):
if (imo+1 >= i):
ans = i-1
print(ans)
main()
``` | output | 1 | 7,479 | 14 | 14,959 |
Provide a correct Python 3 solution for this coding contest problem.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5 | instruction | 0 | 7,480 | 14 | 14,960 |
"Correct Solution:
```
N = int(input())
a=[]
p = 0
for i in range(N):
q=[int(j) for j in input().split()]
a.append([q[0]-1,q[1]])
p = max(p, q[1])
table=[0]*(p+1)
for i in range(len(a)):
table[a[i][0]]+=1
table[a[i][1]]-=1
for i in range(1,p+1):
table[i]+=table[i-1]
ans=0
for i, val in enumerate(table):
if val >= i and i <= N:
ans = max(i, ans)
print(ans)
``` | output | 1 | 7,480 | 14 | 14,961 |
Provide a correct Python 3 solution for this coding contest problem.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5 | instruction | 0 | 7,481 | 14 | 14,962 |
"Correct Solution:
```
N = int(input())
MAX_N = int(1e5) + 2
acc = [0] * (MAX_N + 1)
acc[0] += 1
acc[MAX_N] -= 1
for _ in range(N):
a, b = map(int, input().split())
acc[a] += 1
acc[b + 1] -= 1
for i in range(MAX_N):
acc[i + 1] += acc[i]
for i in reversed(range(MAX_N)):
if acc[i] >= i:
print(i - 1)
break
``` | output | 1 | 7,481 | 14 | 14,963 |
Provide a correct Python 3 solution for this coding contest problem.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5 | instruction | 0 | 7,482 | 14 | 14,964 |
"Correct Solution:
```
n = int(input())
a = [0] * (10**5 + 2)
for i in range(n) :
mi, ma = map(int, input().split())
a[mi-1] += 1
a[ma] -= 1
for i in range(n) :
a[i+1] += a[i]
ans = 0
for i in range(n, 0, -1) :
if a[i] >= i :
ans = i
break
print(ans)
``` | output | 1 | 7,482 | 14 | 14,965 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5
Submitted Solution:
```
N = int(input())
MAXP = 100010
imos = [0 for i in range(MAXP)]
for i in range(N):
a,b = map(int,input().split())
imos[a-1] += 1
imos[b] -= 1
nums = [0]
for im in imos:
nums.append(nums[-1] + im)
for i,p in enumerate(reversed(nums)):
n = MAXP - 1 - i
if n == 0: break
if p >= n:
print(n)
exit()
print(0)
``` | instruction | 0 | 7,483 | 14 | 14,966 |
Yes | output | 1 | 7,483 | 14 | 14,967 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5
Submitted Solution:
```
N = int(input())
R = [0 for i in range(100003)]
for i in range(N):
a, b = map(int, input().split())
R[a] += 1
R[b+1] -= 1
x = 1
y = 1
for i in range(100003):
x += R[i]
if x >= i:
y = max(i, y)
print(y-1)
``` | instruction | 0 | 7,484 | 14 | 14,968 |
Yes | output | 1 | 7,484 | 14 | 14,969 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input())
li = [0] * (10**5 + 3)
for _ in range(n):
a, b = map(int, input().split())
li[a] += 1
li[b + 1] -= 1
for i in range(n + 1):
li[i + 1] += li[i]
ans = 0
for i in range(1, n + 2):
if i - 1 <= li[i]:
ans = max(ans, i - 1)
print(ans)
``` | instruction | 0 | 7,485 | 14 | 14,970 |
Yes | output | 1 | 7,485 | 14 | 14,971 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5
Submitted Solution:
```
#!/usr/bin/env python3
n = int(input())
# imos[k]: γγγγγ‘γγγ©kδΊΊθͺγγ¨γγ«ζ₯γγζ倧人ζ°
imos = [0] * (10**5 + 2)
for _ in range(n):
a, b = map(int, input().split())
imos[a - 1] += 1
imos[b] -= 1
for i in range(n):
imos[i + 1] += imos[i]
ans = 0
for i in range(n + 1):
if imos[i] >= i:
ans = max(ans, i)
print(ans)
``` | instruction | 0 | 7,486 | 14 | 14,972 |
Yes | output | 1 | 7,486 | 14 | 14,973 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5
Submitted Solution:
```
N = int(input())
t = [0 for i in range(100002)]
for i in range(N):
a,b = [int(i) for i in input().split(' ')]
t[a-1] +=1
t[b] -= 1
ans = 0
for i in range(1,100001):
t[i] += t[i-1]
if (t[i] > i):
ans = i
print(ans)
``` | instruction | 0 | 7,487 | 14 | 14,974 |
No | output | 1 | 7,487 | 14 | 14,975 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5
Submitted Solution:
```
N=int(input())
friendlist=[0 for i in range(N+2)]
for i in range(N):
a,b=[int(j) for j in input().split(" ")]
for k in range(a,min(b+1,N+2)):
friendlist[k]+=1
#ab=[[int(i) for i in input().split(" ")] for j in range(N)]
#print(ab)
k=N+1
i=0
while k>=0:
if friendlist[k]>=k-1:
i=k-1
break
k-=1
print(i)
``` | instruction | 0 | 7,488 | 14 | 14,976 |
No | output | 1 | 7,488 | 14 | 14,977 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5
Submitted Solution:
```
N=int(input())
friendlist=[0 for i in range(N+1)]
for i in range(N):
a,b=[int(j) for j in input().split(" ")]
for k in range(a,min(N+1,b+1)):
friendlist[k]+=1
k=N
i=0
while k>=0:
if friendlist[k]>=k-1:
i=k-1
break
k-=1
print(i)
``` | instruction | 0 | 7,489 | 14 | 14,978 |
No | output | 1 | 7,489 | 14 | 14,979 |
Evaluate the correctness of the submitted Python 3 solution to the coding contest problem. Provide a "Yes" or "No" response.
From tomorrow, the long-awaited summer vacation will begin. So I decided to invite my friends to go out to the sea.
However, many of my friends are shy. They would hate it if they knew that too many people would come with them.
Besides, many of my friends want to stand out. They will probably hate it if they know that not many people come with them.
Also, some of my friends are always shy, though they always want to stand out. They will surely hate if too many or too few people come with them.
This kind of thing should be more fun to do in large numbers. That's why I want to invite as many friends as possible. However, it is not good to force a friend you dislike.
How many friends can I invite up to?
I'm not very good at this kind of problem that seems to use my head. So I have a request for you. If you don't mind, could you solve this problem for me? No, I'm not overdoing it. But if I'm slacking off, I'm very happy.
Input
N
a1 b1
a2 b2
..
..
..
aN bN
The integer N (1 β€ N β€ 100,000) is written on the first line of the input. This represents the number of friends.
On the following N lines, the integer ai and the integer bi (2 β€ ai β€ bi β€ 100,001) are written separated by blanks. The integers ai and bi written on the 1 + i line indicate that the i-th friend does not want to go to the sea unless the number of people going to the sea is ai or more and bi or less. Note that the number of people going to the sea includes "I".
Output
Output the maximum number of friends you can invite to the sea so that no friends dislike it.
Examples
Input
4
2 5
4 7
2 4
3 6
Output
3
Input
5
8 100001
7 100001
12 100001
8 100001
3 100001
Output
0
Input
6
2 9
4 8
6 7
6 6
5 7
2 100001
Output
5
Submitted Solution:
```
N = int(input())
P = []
for i in range(N):
a, b = map(int, input().split())
P.append((a, b))
Fmaxim = 0
for i in range(N+1):
Pcount = 1
for a, b in P:
if a <= i+1 <= b:
Pcount += 1
if Pcount >= i+1:
Fmaxim = max(Fmaxim, i)
print(Fmaxim)
``` | instruction | 0 | 7,490 | 14 | 14,980 |
No | output | 1 | 7,490 | 14 | 14,981 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010. | instruction | 0 | 7,650 | 14 | 15,300 |
Tags: brute force, data structures, implementation
Correct Solution:
```
n=int(input())
s=input()
a=[0]*10
for i in range(n):
if s[i]=='L':
for i in range(10):
if a[i]==0:
a[i]=1
break
continue
if s[i]=='R':
for i in range(9,-1,-1):
if a[i]==0:
a[i]=1
break
continue
a[int(s[i])]=0
print("".join(str(e) for e in a))
``` | output | 1 | 7,650 | 14 | 15,301 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010. | instruction | 0 | 7,651 | 14 | 15,302 |
Tags: brute force, data structures, implementation
Correct Solution:
```
n = int(input())
b = []
res = 'a'
for i in range(10):
b.append(0)
l = 0
r = 0
a = input()
for i in range(n):
if a[i] == "L":
k = b.index(0)
b[k] = 1
elif a[i] == "R":
b.reverse()
b[b.index(0)] = 1
b.reverse()
else:
b[int(a[i])] = 0
for i in range(10):
res += str(b[i])
print(res[1:11])
``` | output | 1 | 7,651 | 14 | 15,303 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010. | instruction | 0 | 7,652 | 14 | 15,304 |
Tags: brute force, data structures, implementation
Correct Solution:
```
N = int(input())
events = list(input())
hotel = [False] * 10
for e in events:
if e == 'L':
for i in range(10):
if not hotel[i]: break
hotel[i] = True
elif e == 'R':
for i in range(9, -1, -1):
if not hotel[i]: break
hotel[i] = True
else:
hotel[int(e)] = False
print("".join(map(lambda x: "1" if x else "0", hotel)))
``` | output | 1 | 7,652 | 14 | 15,305 |
Provide tags and a correct Python 3 solution for this coding contest problem.
Amugae has a hotel consisting of 10 rooms. The rooms are numbered from 0 to 9 from left to right.
The hotel has two entrances β one from the left end, and another from the right end. When a customer arrives to the hotel through the left entrance, they are assigned to an empty room closest to the left entrance. Similarly, when a customer arrives at the hotel through the right entrance, they are assigned to an empty room closest to the right entrance.
One day, Amugae lost the room assignment list. Thankfully Amugae's memory is perfect, and he remembers all of the customers: when a customer arrived, from which entrance, and when they left the hotel. Initially the hotel was empty. Write a program that recovers the room assignment list from Amugae's memory.
Input
The first line consists of an integer n (1 β€ n β€ 10^5), the number of events in Amugae's memory.
The second line consists of a string of length n describing the events in chronological order. Each character represents:
* 'L': A customer arrives from the left entrance.
* 'R': A customer arrives from the right entrance.
* '0', '1', ..., '9': The customer in room x (0, 1, ..., 9 respectively) leaves.
It is guaranteed that there is at least one empty room when a customer arrives, and there is a customer in the room x when x (0, 1, ..., 9) is given. Also, all the rooms are initially empty.
Output
In the only line, output the hotel room's assignment status, from room 0 to room 9. Represent an empty room as '0', and an occupied room as '1', without spaces.
Examples
Input
8
LLRL1RL1
Output
1010000011
Input
9
L0L0LLRR9
Output
1100000010
Note
In the first example, hotel room's assignment status after each action is as follows.
* First of all, all rooms are empty. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* L: one more customer from the left entrance. Assignment status is 1110000001.
* 1: the customer in room 1 leaves. Assignment status is 1010000001.
* R: one more customer from the right entrance. Assignment status is 1010000011.
* L: one more customer from the left entrance. Assignment status is 1110000011.
* 1: the customer in room 1 leaves. Assignment status is 1010000011.
So after all, hotel room's final assignment status is 1010000011.
In the second example, hotel room's assignment status after each action is as follows.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000 again.
* 0: the customer in room 0 leaves. Assignment status is 0000000000.
* L: a customer arrives to the hotel through the left entrance. Assignment status is 1000000000.
* L: one more customer from the left entrance. Assignment status is 1100000000.
* R: one more customer from the right entrance. Assignment status is 1100000001.
* R: one more customer from the right entrance. Assignment status is 1100000011.
* 9: the customer in room 9 leaves. Assignment status is 1100000010.
So after all, hotel room's final assignment status is 1100000010. | instruction | 0 | 7,653 | 14 | 15,306 |
Tags: brute force, data structures, implementation
Correct Solution:
```
n = int(input())
ans = [0]*10
a = list(input())
lastl=0
lastr=9
for i in range (0,len(a)):
lastl=0
lastr=9
if a[i]=='L':
while ans[lastl]== 1:
lastl += 1
ans[lastl]=1
elif a[i] == 'R':
while ans[lastr]== 1:
lastr -= 1
ans[lastr]=1
elif int(int(a[i])) <=4:
ans[int(a[i])] =0
elif int(a[i]) >=5:
ans[int(a[i])] =0
for i in range (0,10):
print(ans[i],end="")
``` | output | 1 | 7,653 | 14 | 15,307 |
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